Kanada 2011

Assume the contrary: there exist $a$ and $b$ of the prescribed form, such that $b\ge a$ and $a$ divides $b$. Then $a$ divides $b-a$.

Claim: $a$ is not divisible by $3$ but $b-a$ is divisible by $9$. Indeed, the sum of the digits is $10(1+\cdots +7)=280$, for both $a$ and $b$. [Here one needs to know or prove that an integer $n$ is equivalent to the sum of its digits modulo $3$ and modulo $9$.]

We conclude that $b-a$ is divisible by $9a$. But this is impossible, since $9a$ has $71$ digits and $b$ has only $70$ digits, so $9a>b>b-a$. $\square$